Question

\[\int\limits_{1}^{3} e^{3x} dx\]

Answer 1

start with substituting 3x = u
du =.... ?

Answer 2

@Jlockwo3 can you find du=.... ?

Answer 3

is it 1/3 du

Answer 4

i was asking du in terms of dx
u= 3x
differentiate, w.r.t x., what u get ?

Answer 5

guess idk what the question is, im lost

Answer 6

but you can differentiate, right ?
what you get after differentiating u= 3x ?

Answer 7

i did e^3x = ln 3x

Answer 8

e^3x doesn't equal ln 3x....
before starting integration, we must know basic derivatives ?
if you forgot, please revise the derivatives formulas!

Answer 9

u = 3x
1/3du=dx

Answer 10

3x just equals 3 with derivatives

Answer 11

but idk where to take this problem

Answer 12

1/3 du = dx is correct :)

Answer 13

so, we should change our integration variable from x to u

Answer 14

ok so then e^u * 1/3 du

Answer 15

yes! good :)
now change the limits too!
when x= 1, u=... ?
when x=3, u=.... ?

Answer 16

e^1 = 2.71
e^3 = 20.09

Answer 17

........ 3x = u, so x is just u/3 !
right ?
now try again :)

Answer 18

ok 1/3 = .333 3/3 = 1

Answer 19

correct, so now we have
\(\large \int \limits_{1/3}^1e^udu/3=\large (1/3)\int \limits_{1/3}^1e^udu\)
can you integrate e^u ??

Answer 20

\[\int\limits_{1}^{3}e ^{3x}dx = e ^{u} = e ^{u}* 1/3 du \]

Answer 21

constants can be taken out of integration :)

Answer 22

= e^3x * 1/3 = e^3*1 *1/3 - e^3*3 * 1/3

Answer 23

2694.33

Answer 24

i did that backwards, meant to put the 3 - 1 not 1-3

Answer 25

i don't understand what u did , there shouldn't be any x, once we changed the variable from x to u, there should just be 'u'
\(\large \int \limits_{1/3}^1e^udu/3=\large (1/3)\int \limits_{1/3}^1e^udu=(1/3)[e^u]^1_{1/3}\)
because integral of e^u du is just e^u

Answer 26

\(\large (1/3) [e^1 -e^{1/3}]\)
^thats substituting the limits
we can keep this as final answer or use calculator and find decimal approximation...

Answer 27

wow this was hard....thank you

Answer 28

with some practice, it'll become easy :)
welcome ^_^